Problem 74 - Chemistry

Question

Consider the reaction $$2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)$$ At \(430^{\circ} \mathrm{C},\) an equilibrium mixture consists of 0.020 mole of \(\mathrm{O}_{2}, 0.040\) mole of \(\mathrm{NO},\) and 0.96 mole of \(\mathrm{NO}_{2}\). Calculate \(K_{P}\) for the reaction, given that the total pressure is 0.20 atm.

Step-by-Step Solution

Answer

The equilibrium constant \(K_P\) for the reaction is \(1.1 \times 10^3\).

Step 1: Calculate partial pressures

First, the partial pressures of each gas at equilibrium need to be calculated. The partial pressure of a gas in a mixture is calculated by multiplying the total pressure by the mole fraction of the gas. The mole fraction of a gas is its number of moles divided by the total number of moles in the mixture. The total number of moles in the mixture is \(0.020 \, mol \, O_2 + 0.040 \, mol \, NO + 0.96 \, mol \, NO_2 = 1.02 \, mol\). So the partial pressures of \(O_2\), \(NO\), and \(NO_2\) are \(P_{O_2} = 0.020/1.02 \times 0.20 \, atm = 0.00392 \, atm\), \(P_{NO} = 0.040/1.02 \times 0.20 \, atm = 0.00784 \, atm\), and \(P_{NO_2} = 0.96/1.02 \times 0.20 \, atm = 0.188 \, atm\).

Step 2: Apply the equilibrium constant formula

The equilibrium constant \(K_P\) is given by: \(K_P = \dfrac {{(P_{NO_2})^2}}{{P_{NO}^2 \cdot P_{O_2}}}\) where \(P_{NO}\), \(P_{O_2}\), and \(P_{NO_2}\) are the equilibrium partial pressures of the gases and the exponents represent the stoichiometric coefficients in the balanced chemical equation.

Step 3: Calculate \(K_P\)

Substitute the values obtained in step 1 into the equilibrium constant formula from step 2: \(K_P = \dfrac {{(0.188)^2}}{{(0.00784)^2 \cdot 0.00392}}\). Solving this gives \(K_P = 1.1 \times 10^3\).

Key Concepts

Equilibrium Partial Pressures

To understand chemical reactions that reach a state of balance, or equilibrium, we often look at the equilibrium partial pressures of the gases involved in the reaction. The partial pressure is defined as the pressure exerted by a single gas in a mixture of gases. It's an important concept because it reflects the concentration of that particular gas in the mixture.

When we deal with equilibrium systems, the partial pressure plays a pivotal role in determining the position of equilibrium, directly affecting the reaction's equilibrium constant, denoted as KP. In practice, we calculate a gas's partial pressure by multiplying the total pressure of the gas mixture by the gas's mole fraction. For instance, if a gas contributes 30% of the moles in a mixture and the total pressure is 1 atmosphere, the gas's partial pressure would be 0.3 atmospheres. This type of calculation is essential for understanding how changes in pressure and concentration influence the direction and extent of a chemical reaction at equilibrium.

Equilibrium Constant Formula KP

The equilibrium constant formula KP provides a way to quantitatively describe the position of equilibrium for a gaseous reaction. It's expressed in terms of the partial pressures of the gases involved and relates to the reaction's balanced chemical equation. Depending on the stoichiometry of the reaction, the formula will look different as the exponents correspond to the coefficients in the balanced equation.

For a general reaction where gases A and B react to form gases C and D, the KP formula might look like this: \( KP = \frac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b} \), where P represents the partial pressure, and the lower-case letters represent the stoichiometric coefficients from the balanced equation. An important nuance to remember is that KP is dimensionless only if the number of moles of gaseous reactants is equal to the number of moles of gaseous products, which often isn't the case. That's why units like atm can sometimes be included, implying that the reaction involves a change in the number of moles of gas.

Stoichiometry

Stoichiometry is the backbone of chemistry that deals with the quantitative relationships of reactants and products in a chemical reaction. It involves using the balanced chemical equation to determine the proportions of substances needed or produced. In the equation \(2 NO(g) + O_2(g) \rightleftharpoons 2 NO_2(g)\), for instance, stoichiometry tells us that two moles of NO gas react with one mole of O2 gas to produce two moles of NO2 gas.

In the context of equilibrium constant expressions, stoichiometry plays a critical role as the coefficients in the balanced equation become the exponents in the expression for KP. A sound understanding of stoichiometry allows students to manipulate and understand reaction conditions such as concentration changes or volume changes, which can shift the equilibrium position.

Mole Fraction

The mole fraction is a dimensionless quantity expressing the ratio of the number of moles of a component to the total number of moles in a mixture. It's a way to describe the composition of a mixture and is crucial in the context of partial pressures and eventually the equilibrium constant KP.

As a simple ratio, the mole fraction has the advantage of being a straightforward and temperature-independent measure of concentration. This makes mole fractions particularly useful in gas laws where temperature plays a role in determining the behavior of gases. To calculate the mole fraction, one divides the number of moles of one substance by the total number of moles in the mixture. This can be represented as: \( X_i = \frac{n_i}{n_{total}} \), where \(X_i\) is the mole fraction of substance \(i\), \(n_i\) is the number of moles of the substance, and \(n_{total}\) is the total number of moles in the mixture.